Duality of A-hypergeometric systems

Uli Walther Purdue University

walther@math.purdue.edu November 2004 Duality

D_{n} = ℂ[x_{1}, ∂_{1}, …, x_{n}, ∂_{n}], M a finite holonomic left D_{n}-module, F^{•} → → M a D_{n}-free resolution.

τ : D_{n} → D_{n}, x^{u}∂^{v} → (−∂)^{v}x^{u} is an equivalence

D_{n} − mods ≅ mods −D_{n}.

Definition: The holonomic dual 𝔻(M) of M is

τ (Ext _{D_{n}}^{n}(M, D_{n})) = τ (H^{n}(Hom _{D_{n}}(F^{•}, D_{n})))

Note: Ext _{D_{n}}^{i}(M, D_{n}) ∈ mods −D_{n}, so 𝔻(M) ∈ D_{n} − mods .

Remark Sol : Hol_{n} → Loc-Const _{n},

Sol(M) = ℝ Hom _{D_{n}^{an}}(M^{an}, O_{n}). One has

Sol(𝔻(M)) ≅ Verd (Sol(M)).

Question: (Takayama) If M = H_{A}(β), what is 𝔻(M)? In particular, is 𝔻(H_{A}(β)) an A-hypergeometric system?

Example: A = ( \begin{matrix}1 & 1 & 1 \\ 0 & 1 & 2 \end{matrix} ).

E_{1} = x_{1}∂_{1} + x_{2}∂_{2} + x_{3}∂_{3}, E_{2} = x_{2}∂_{2} + 2x_{3}∂_{3}. A resolution of H_{A}(β):

0 → D_{n} ⟶^{·ϕ_{3}} D_{n}^{3} ⟶^{·ϕ_{2}} D_{n}^{3} ⟶^{·ϕ_{1}} D_{n} → H_{A}(β) → 0

with maps

ϕ_{1} = ( D_{2}^{2} − D_{1}D_{3} E_{1} − β_{1} E_{2} − β_{2} )^{T}

ϕ_{2} = ( \begin{matrix}E_{2} − β_{2} + 2 & E_{1} − β_{1} + 2 & 0 \\ 0 & −D_{2}^{2} + D_{1}D_{3} & −E_{2} + β_{2} \\ −D_{2}^{2} + D_{1}D_{3} & 0 & E_{1} − β_{1} \end{matrix} )^{T}

ϕ^{3} = ( \begin{matrix}−E_{1} + β_{1} + −2 \\ E_{2} − β_{2} + 2 \\ −D_{2}^{2} + D_{1}D_{3} \end{matrix} )^{T}

Hence the dual is

τ (\frac{D_{n}}{(−E_{1} + β_{2} − 2, E_{2} − β_{2} + 2, −D_{2}^{2} + D_{1}D_{3})D_{n}} )

= H_{A}(−β − \binom{1}{1}).

Another look at the resolution:

D_{n}

[figure: D_{n}; ·(−E_{2}+β_{2}); ⊕; ·(E_{1}−β_{1}); D_{n}]

[figure: ·(E_{1}−β_{1}); D_{n}; ·(E_{2}−β_{2})]

D_{n}^{·(∂_{2}^{2}−∂_{1}∂_{2}) }

[figure: D_{n}; ·(∂_{2}^{2}−∂_{1}∂_{2}); ·(−E_{2}+β_{2}−2); ⊕; ·(E_{1}−β_{1}+2); D_{n}; ·(∂_{2}^{2}−∂_{1}∂_{2})]

[figure: ·(E_{1}−β_{1}+2); D_{n}; ·(E_{2}−β_{2}+2)]

and Hom _{D_{n}}(F^{•}, D_{n}):

D_{n}
[figure: D_{n}; (−E_{2}+β_{2})·; ⊕; (E_{1}−β_{1})·; D_{n}]
[figure: (E_{1}−β_{1})·; D_{n}; (E_{2}−β_{2})·]

D_{n}^{(∂_{2}^{2}−∂_{1}∂_{2})·_{}}
[figure: D_{n}; (−E_{2}+β_{2}−2)·; ⊕; (E_{1}−β_{1}+2)·; D_{n}]
[figure: (E_{1}−β_{1}+2)·; D_{n}; (E_{2}−β_{2}+2)·]

These are right modules.

Let ε_{A} = ∑ a_{i}.

Proposition: If I_{A} is a Gorenstein ideal

(i.e., F^{•} → → R_{n}/I_{A} is symmetric of length d) then 𝔻(H_{A}(β)) = H_{A}(−β − ε_{A} + c_{A}) where c_{A} is the canonical degree of R_{n}/I_{A}.

Problem: Gorenstein not too frequent. Good cases: n = d + 1, and some nice A.

What is the problem?

Proposition: If I_{A} is Cohen-Macaulay, then

𝔻(H_{A}(β)) = H_{0}(Ext _{R_{n}}^{n−d}(S_{A}, R_{n}), −β − ε_{A}).

But C_{A} := Ext _{R_{n}}^{n−d}(S_{A}, R_{n}) ≅ R̸_{n}/I_{A}.

Observation: There is

0 → S_{A}(−c_{A}) ↪ C_{A} → → N → 0 (#) with dim (N) < d. Corollary: If A is CM and β is generic, then

𝔻(H_{A}(β)) ≅ H_{A}(γ)

for a suitable γ. Basic idea: (#) gives

⋯
H_{1}(S_{A}, γ + c_{A}) → H_{1}(C_{A}, γ + c_{A}) → H_{1}(N, γ + c_{A}) →
H_{0}(S_{A}, γ + c_{A}) → H_{0}(C_{A}, γ + c_{A}) → H_{0}(N, γ + c_{A}) → 0

By [MMW], if −β ∉ qdeg (N) the N-column vanishes. So, 𝔻H_{A}(β) = 𝔻H_{0}(S_{A}, β) equals

H_{0}(C_{A}, −β − ε_{A}) = H_{0}(S_{A}, −β − ε_{A} + c_{A})

What about non-CM systems? Theorem For all A, for generic β,

𝔻(H_{A}(β)) ≅ H_{A}(γ)

is a GKZ-system. Idea:

1. For i < n − d and most β, −β ∉ qdeg (Ext _{R_{n}}^{i}(S_{A}, R_{n})).

2. Hence for most β, 𝔻(H_{A}(β)) = H_{0}(Ext _{R_{n}}^{n−d}(S_{A}, R_{n}), −β − ε_{A}).

3. 0 → S_{A}(−c) → Ext _{R_{n}}^{n−d}(S_{A}, R_{n}) → N → 0.

4. For most β, H_{0}(Ext _{R_{n}}^{n−d}(S_{A}, R_{n}), −β−ε_{A}) ≅ H_{A}(−β−ε_{A}+c).

Example: A = ( \begin{matrix}1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 4 \end{matrix} ).

As before, C_{A} = Ext _{R_{n}}^{4−2} (S_{A}, R_{n}). Put N = Ext _{R_{n}}^{4−1} (S_{A}, R_{n}) = ℂ \binom{−5}{−10}.

Let γ = −β − ε_{A} = −β − ( \begin{matrix}4 \\ 8 \end{matrix} ).

There is a spectral sequence with E_{2}-term

[figure: 0  0  0  0  0; 0 H_{0}(C_{A}, γ)  H_{0}(N, γ)  0  0; 0; H_{1}(C_{A}, γ); d_{2}; H_{0}(N, γ); 0; 0; 0 H_{2}(C_{A}, γ)  H_{0}(N, γ)  0  0; 0  0  0  0  0]

with abutment 𝔻(H_{A}(β)).

Consider β = ( \begin{matrix}1 \\ 2 \end{matrix} ), γ = ( \begin{matrix}−5 \\ −10 \end{matrix} ).

Since N = ℂ ( \begin{matrix}−5 \\ −10 \end{matrix} ), H_{i}(N, γ) = O_{4}⊗_{ℤ}⋀^{i}(ℤ^{2}).

Since C_{A} is C-M, H_{i}(C_{A}, −) = 0 for i > 0. So, the E^{2}-page is

[figure: 0  0  0  0  0; 0 H_{0}(C_{A}, γ)  O_{4}  0  0; 0; 0; d_{2}; (O_{4})^{2}; 0; 0; 0  0  O_{4}  0  0; 0  0  0  0  0]

We get the exact sequence 0 → O_{4} → H_{0}(C, γ) → 𝔻(H_{A}(β)) → (O_{4})^{2} → 0

Contradiction!

Question: Which β fail to give dual GKZ’s?